finite elements 2
DESCRIPTION
FEATRANSCRIPT
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One-Dimensional Finite ElementAnalysis for a Continuum System
Computer Modelling Techniques
(Module ode: MM3CMT)
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Governing equation for a tensioned string:
2
20
d wT f x
dx
w Deflection of the stringx Longitudinal coordinateT Tension in the stringf Applied transverse force
(force per unit length)
Boundary conditions:
0 or 0
at andl r
dww
dx
x x x
T
T
w=w(x)
f=f(x)z, w
x
2D counterpart:
1-D form of the Poisson equation
2 2
2 2,
w wf x y
x y
Governs physical problems, such asmembranes, torsion of prismatic bars,steady state heat transfer, etc.
Derivation of the governing equation:
2
20
d wT f x
dx
Small deflection assumptionTension not affected by deflection
2
20
dw dw d wT T dx f x dx
dx dx dx
Vertical equilibrium:
2
2
( )
( ) ( )
dww x dx w x dx
dx
dw x dx dw x d w xdx
dx dx dx
TT
w=w(x)
f=f(x)z, w
x
f(x)dx
dw(x)/dxdw(x+dx)/dx=dw/dx+(d2w/dx2)dxT
T
dxx
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Solutions to differential equations (Boundary value problems)
2
20
d wT f x
dx
Solution: w=w(x) a function of x over the interval [xl , xr]
The solution satisfies:(1) the governing equation within the interval(2) the boundary conditions at the ends of the interval
Boundary conditions:
0 at andl rw x x x
Governing equation:
within (xl , xr)
[Note]: The other type of boundary condition dw/dx=0 is of a differentmathematical nature and will not be considered at this stage of development
Example: [xl , xr]=[0, L] and
The solution:
sinx
f x AL
2
2sin
L A xw x
T L
• Discretise the domain into finite number of elements
• Construct an approximate field using a polynomial within each element
1
1
( ) i ii i
i i
w ww x w x x
x x
x1 x2 x3 x4 xi-1 xi xi+1 xn xn+1
w1w2
w4w3
wi+1wiwi-1
wn+1wnith element ith node
ith d.o.f.
Mesh:
Interpolation:
FE Approximation
11 1 1
1 1
( ) i ii i i i i i
i i i i
x x x xw x w w N x w N x w
x x x x
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2
2( ) 0
d wT f x
dx weak form
Formulation1
1
2
2( ) ( ) 0
nx
j
x
d wT f x x dx
dx
11 1
11 1
2
2( ) ( ) ( ) ( ) 0
nn nxx xj
j j j
xx x
dd w dw dwT f x x dx T T f x x dx
dx dx dx dx
Integration by parts:
1 1
11
( ) ( ) ( ) ( ) 0n i
i
x xnj j
j jix x
d ddw dwT f x x dx T f x x dx
dx dx dx dx
1
1
0nx
j
x
dwT
dx
Boundary conditions to be considered later
Governing equation
Integration by parts: r r
r
l
l l
x xx
xx x
du dvvdx uv u dx
dx dx
Let & j
dwu v
dx
11
1
1i i i
i
i i
i
i
ewN x N x B
w
dwN x w N x w
dx
Interpolation:
11 1
1
1
1
1 1
( )
ei
i
i ii i i i i i
i i i i
i
i
x x x xw x w w N x w N x w
x x x x
wN x N x S
w
1 1
1
( ) i i i i
i
i
fSf x N x f N x f
f
1
1 1
1 1i i
i i i i
N x N xx x x x
11
1 1
i ii i
i i i i
x x x xN x N x
x x x x
Shape functions:• A shape function is associated witheach node• Shape functions share the same formfor all elements of the same type• More universal forms are expressedusing nondimensionalised coordinates
1
e i
i
w
w
Shape functions
Kinematic matrix
1 1
( )
( )
i i
i i
f f x
f f x
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Formulation of an element
edw
Bdx
1
( )i
i
xe
i i
x
N T B N f x dx
For the ith element:
1
1 1 ( )i
i
xe
i i
x
N T B N f x dx
1
1
( )i
i
xT e i
ix
NB T B f x dx
N
e e e
K F
1
1 1
1 1
( )
i
i
i i
i i
xe T
x
x xTe i i
i ix x
K B T B dx
N fF f x dx S S dx
N f
1 1
1
( ) i
i i i i
i
ff x N x f N x f S
f
1 1
( ) ( ) ( ) ( )i i
i i
x x
j j
j j
x x
d dNdw dwT f x x dx T N x f x dx
dx dx dx dx
( ) ( )j jx N x
j=i, i+1 for the ith element
Let
where
1
2 21 1 1 1
1
1
1 1
1
1
1
1
1
1
1
1 1 11 11 1
1 1 1 1
i i
i i i i i i i i
i i
i
i i i i
i i i i
i
i i
i
i
i i
i
i i
x x T TT
x x x x x x x x
x x
x x
x x x
NT N N
N
NN
x x x
x x x x
x
N xN
x
2
1 1
2 21 1 1
1 i i i
i i i i i i
x x x x x x
x x x x x x x x x
1
1 1
1 1i i
i i i i
N x N xx x x x
11
1 1
i ii i
i i i i
x x x xN x N x
x x x x
11
1 11
11
2 3 3
1 1 1
2 2 3
1 1 1 1
2 3 3
1
1 1
3 3
1 1 1
2 2 6
1 1
3 3
ii
ii
i ii
ii i
ii
ii
xx
i i i ix
x
x xx
i i i i i i ix
x x
xx
i i i ix
x
x x dx x x x x
x x x x dx x x x x x x dx x x
x x dx x x x x
1 1
1 1
1
1 1
1
1 1 1 1
1
1
1 1
1 1
2 1
1 26
i i
i i
i i
i i
i
i i
i
i
i
x xTe
i ix x
x xTe i i i ii i
i i i ix
i
ix
TK B T B dx dx
x x
F f f fx xF S S dx dx
F f f
NT N N
N
NN N
N f
11
1
2
26
i ii i
i i
f fx x
f f
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Assembly from elements to structure
ith element:
1 1( ) ( ) ( ) 1 1
11 1
1 1
12
6
12
6
i i i ii i i i i i i i
ii i i i
i i i i
T Tf f x x
x x x x wK F
wT Tf f x x
x x x x
( ) ( ) ( )
1
0n
i i i
i
K F K F
Structure:
Full length of the string L n equal length element:
Uniformly distributed load:
( ) ( ) ( )
1
1 1 1
1 1 12
i i i i
i
wnT LfK F
wL n
1i if f f
1i i
Lx x
n
1,1 1,2 1
2,1 2,3
3,2 3,3 3,4 3
4,3 4,4 4
1, 1 1, 1
, 1
2,2
,
2
, 1
1, 1, 1 1
0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0
n n n n n
n n n n n n n
n n n n n
K K w
K K
K K K w
K K w
K K w
K K K w
w
K w
K
K
1
3
4
1
1
2
0
n
n
n
F
F
F
F
F
F
F
(1) (1) (1)11,1 1,2 1
(1) (12
) (1)2,1 2,2 2
wK K F
K K Fw
(2) (2) (2)1,1 1,2 1
(2) (2) (2)32,1 2,2 2
2K K F
wK K F
w
K F
(1) (2)2,2 2,2 1,1
(1) (2)2 2 1
K K K
F F F
Assembly: effects and implication
The same applies to K3,3 , …, Kn,n due to the continuity at Nodes 3, …, n.
Continuity at Node 2
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K F
[K]: Structural stiffness matrix{F}: Structural load vector{}: Structural degrees of freedom vector
Numerical features of the structural stiffness matrix:1) Symmetric 2) Banded (half band width=2) 3) Semi positive definite (singular)
2 1 2 1
2 1 2 1 3 2 3 2
3 2 3 2 4 3 4 3
4 3 4 3 5 4
1 2 1 1
1 1 1 1
0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0
n n n n n n
n n n n n n n n
T T
x x x x
T T T T
x x x x x x x x
T T T T
x x x x x x x x
T T T
x x x x x x
T T T
x x x x x x
T T T T
x x x x x x x x
T
x
1 1 2 2 1
2 1 2 2 1 2 3 3 2
3 2 3 3 2 3 4 4 3
4 3 4 4 3 4 5 5 4
1 2 1
1
1 1
2
2 2
2 2
2 21
6
2n n n
n
n
n n n n
w f f x x
w f f x x f f x x
w f f x x f f x x
w f f x x f f x x
w f f x
w
w
T
x x x
1 2 1 1
1 1 1 1
1 1
2
2 2
2
n n n n n n
n n n n n n n n
n n n n
x f f x x
f f x x f f x x
f f x x
Stiffness equation:
Singularity: Rigid body motions (translation & rotation)
1
2
3
4
1
1
1 1 0 0 0 0 0 1
1 2 1 0 0 0 0 2
0 1 2 1 0 0 0 2
0 0 1 2 0 0 0 2
2
0 0 0 0 2 1 0 2
0 0 0 0 1 2 1 2
0 0 0 0 0 1 1 1
n
n
n
w
w
w
wnT Lf
L n
w
w
w
n equal length element &uniformly distributed load
Equivalence:
2
20
d wT f x K F
dx
A differential equation has been converted into a set ofsimultaneous algebraic equations
Algebraic equations can be solved effectively with computers
CMT is to interpret physical problems into a form which canbe easily dealt with using computers
Typical approaches employ a mesh of one form or another,e.g. FE, FD and FV, with FE being particularly popular forsolids and structures
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Boundary conditions:
2
20 or
d wT f x K F
dx
As they are, both have infinite number of solutions
In either case, the solution becomes unique after imposition of boundary conditions
1 1 1 10 0 &l r
n nx x x xw w w w R R
Nodes: 1 2 3 4 i i+1 n-1 n n+1
Elements: 1 2 3 i n-1 n
(Reactions)
Governing equation
1 1 0nw w
Nodes: 1 2 3 4 i i+1 n-1 n n+1
Elements: 1 2 3 i n-1 n
(Reactions)
K F
After imposition of appropriate boundary conditions (shown as underscored when necessary):
KSymmetricBandedPositive definite
Significance: Guarantee the existence anduniqueness of the solution; minimise thedemand on memory required for solving theproblem and allows the most effective solutiontechnique to be used
2 1
2 1 2 1 3 2 3 2
3 2 3 2 4 3 4 3
4 3 4 3 5 4
1 2 1 1
50
1 1 1 1
0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0
10
n n n n n n
n n n n n n n n
n
T
x x
T T T T
x x x x x x x x
T T T T
x x x x x x x x
T T T
x x x x x x
T T T
x x x x x x
T T T T
x x x x x x x x
T
x
1 2 2 1
2 1 2 2 1 2 3 3 2
3 2 3 3 2 3 4 4 3
4 3 4 4 3 4 5 5 4
1 2 1 1
1
50
1
1 2
2 2
2 2
2 21
6
2
10
n n n
n
n
n n
n
f f x x
w f f x x f f x x
w f f x x f f x x
w f f x x f f x x
w f f x x
w
w
w
x
2 1 1
1 1 1 1
1 1
2
2 2
2
n n n n
n n n n n n n n
n n n n
f f x x
f f x x f f x x
f f x x
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Forms of solutions:
2
20
d wT f K F
dx
Numerical solution within the ith element: Interpolation
1 10 0l r
nx x x xw w w w
Nodes: 1 2 3 4 i i+1 n-1 n n+1
w1w2
w3
w4
wiwi+1
wn-1
wn+1
wn
Analytical solution: Numerical solution:w=w(x) w1, w2, w3, ……, wn, wn+1
w=w(x)
11 1
11 1
( ) ii ii i i i
ii i i i
wx x x xw x w w N x N x
wx x x x
Example: n=2 & f=0 but concentrated force F at node 2
Solution: 2
2 1 3 2
T Tw P
x x x x
1
2
2 1 3 2
1 1 Pw
x x x x T
Nodes: 1 2 3
w1
w2 w3
P
1 3
1
2R R P
Reactions: 3 21 2
2 1 3 1
x xTR w P
x x x x
2 1
3 2
3 2 3 1
x xTR w P
x x x x
2
4k
TP w
L
If the length of the string is L and node 2 is the midpoint:
2 1 2 11 1
2
2 1 2 1 3 2 3 2
3 3
3 2 3 2
0
0
T T
x x x xw R
T T T Tw P
x x x x x x x xw R
T T
x x x x
Spring
2 11 1
2
2 1 2 1 3 2 3 2
3 3
3 2
0
0
T
x xw R
T T T Tw P
x x x x x x x xw R
T
x x
Imposing constraints:
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Exercises (How do you get there?)
Assuming gravity force, i.e. f= fi= gA, solve the problem with xl=0 & xr=L
1) Number of nodes n+1=3, i.e. with n=2 elements of equal length2) Number of nodes n+1=4, i.e. with n=3 elements of equal length3) Number of nodes n+1=5, i.e. with n=4 elements of equal length
=7.8kg/m g=9.8m/s2 A=10-6m2 L=150m T=100N
2
3
4
1
2 1 0 0 0 1
1 2 1 0 0 1
0 1 2 0 0 1
0 0 0 2 1 1
0 0 0 1 2 1
n
n
w
w
wTn gAL
L n
w
w
Maximum deflection:
n=2 2.151mn=3 1.912mn=4 2.151mn=5 2.065mn=6 2.151mn=7 2.107m
Convergence !
Summary
Continuum systems: 1-D form
Differential equations (boundary value problems) Weak form
Finite element method (FEM)
Basic / new concepts:
Shape functions
Kinematic matrix
Element stiffness matrix
Structural stiffness matrix (assembling process)
Constraints (boundary conditions) & Reactions
Illustration through a tensioned string
2 2
2 2,
w wf x y
x y
2
2( ) 0
d wT f x
dx
2-D:
1-D:
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Tutorial Questions:
1) How are shape functions used in interpolation of continuous functions within an element?2) What is the kinematic matrix [B]? What is the significance of it in FE formulation?3) How is [K]e related to the kinematic matrix [B]?4) Show that shape functions satisfy Ni(xj)=1 if j=i or Ni(xj)=0 if ji.5) What is the physical basis of assembling elements into a structure?6) Prove that fixed boundary conditions at given degrees of freedom can be imposed by setting thecorresponding diagonal components in the structural stiffness matrix to extremely large values.7) Carry out complete analyses (step by step) of a tensioned string with details as given in the examplebut under a transverse load f(x)=sin(x/L), neglecting gravity, using 2 and 4 elements, respectively.8) Having found nodal displacements, how to determine the displacement at a specific point within agiven element?
Multiple Choice Questions (taster):
1) [K]e is singular because (a) it is a matrix; (b) it is symmetric; (c) the finite element is free todisplace as a rigid body; (d) the element store an amount of energy.2) The physical consideration behind the way [K]e is assembled into [K] is (a) continuity betweenneighbouring elements; (b) constant tension among neighbouring elements; (c) presence of theboundary conditions; (d) presence of the load applied.3) Boundary conditions are essential in FE analyses because (a) they simplify the solution of theproblem; (b) they ensure the existence of the solution; (c) they ensure the uniqueness of the solution;(d) I have been told to impose them.